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> 已知:m,n是两个连续自然数(m<n),且q=mn.设p=q+n+q-m,则p()A.总是奇数B.总是偶数C.有时是奇数,有时是偶数D.有时是有理数,有时是无理数-数学
已知:m,n是两个连续自然数(m<n),且q=mn.设p=q+n+q-m,则p()A.总是奇数B.总是偶数C.有时是奇数,有时是偶数D.有时是有理数,有时是无理数-数学
题目简介
已知:m,n是两个连续自然数(m<n),且q=mn.设p=q+n+q-m,则p()A.总是奇数B.总是偶数C.有时是奇数,有时是偶数D.有时是有理数,有时是无理数-数学
题目详情
已知:m,n是两个连续自然数(m<n),且q=mn.设
p=
q+n
+
q-m
,则p( )
A.总是奇数
B.总是偶数
C.有时是奇数,有时是偶数
D.有时是有理数,有时是无理数
题型:单选题
难度:中档
来源:连云港
答案
m、n是两个连续自然数(m<n),则n=m+1,
∵q=mn,
∴q=m(m+1),
∴q+n=m(m+1)+m+1=(m+1)2,q-m=m(m+1)-m=m2,
∴
p=
q+n
+
q-m
=m+1+m=2m+1,
即p的值总是奇数.
故选A.
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题目简介
已知:m,n是两个连续自然数(m<n),且q=mn.设p=q+n+q-m,则p()A.总是奇数B.总是偶数C.有时是奇数,有时是偶数D.有时是有理数,有时是无理数-数学
题目详情
答案
∵q=mn,
∴q=m(m+1),
∴q+n=m(m+1)+m+1=(m+1)2,q-m=m(m+1)-m=m2,
∴p=
即p的值总是奇数.
故选A.